Finding a Basis - Linear Algebra

1. Consider the matrix $A = \begin{pmatrix} 1 & 3 & 5 & -1 \\ -1 & -2 & -5 & 4 \\ 0 & 1 & 0 & 3 \end{pmatrix}$. Which of the following sets forms a basis for the column space of $A$?
   Answer: $\{\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix}\}$ To find a basis for the column space, we identify the pivot columns of the matrix $A$. We row-reduce $A$: $\begin{pmatrix} 1 & 3 & 5 & -1 \\ -1 & -2 & -5 & 4 \\ 0 & 1 & 0 & 3 \end{pmatrix} \xrightarrow{R_2+R_1} \begin{pmatrix} 1 & 3 & 5 & -1 \\ 0 & 1 & 0 & 3 \\ 0 & 1 & 0 & 3 \end{pmatrix} \xrightarrow{R_3-R_2} \begin{pmatrix} 1 & 3 & 5 & -1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ The pivot positions are in the first and second columns. A basis for the column space is formed by the corresponding columns from the original matrix $A$. Thus, a basis is $\{ \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \}$. Choice A incorrectly uses columns from the echelon form. Choice B incorrectly uses rows. Choice D includes all columns, which is a spanning set but is linearly dependent.
2. Let $W$ be the subspace of $\mathbb{R}^3$ spanned by $\{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$, where $\vec{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$, $\vec{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$, and $\vec{v}_3 = \begin{pmatrix} 2 \\ -3 \\ -7 \end{pmatrix}$. Which statement accurately describes a basis for $W$?
   Answer: A basis for $W$ is $\{\vec{v}_1, \vec{v}_2\}$ because $\vec{v}_3$ is a linear combination of $\vec{v}_1$ and $\vec{v}_2$. To determine the basis, we must check for linear dependence. We can see if $\vec{v}_3$ is a linear combination of $\vec{v}_1$ and $\vec{v}_2$ by solving $c_1\vec{v}_1 + c_2\vec{v}_2 = \vec{v}_3$. This gives the system of equations: $c_1 = 2$, $c_2 = -3$, and $-2c_1 + c_2 = -7$. Substituting the first two equations into the third gives $-2(2) + (-3) = -4 - 3 = -7$. The equation holds. Since $\vec{v}_3 = 2\vec{v}_1 - 3\vec{v}_2$, $\vec{v}_3$ is redundant. The remaining vectors, $\vec{v}_1$ and $\vec{v}_2$, are not scalar multiples, so they are linearly independent and form a basis for $W$. Thus, $W$ is a plane in $\mathbb{R}^3$.
3. A student is finding a basis for the column space of a $4 \times 5$ matrix $A$. Which of the following procedures is guaranteed to produce a correct basis?
   Answer: Select the columns of $A$ that correspond to the pivot columns in the reduced row echelon form of $A$. The standard algorithm for finding a basis for the column space (Col $A$) is to row-reduce $A$ to an echelon form, identify the columns that contain pivots, and then select the corresponding columns from the original matrix $A$. Choice A correctly describes this procedure. Choice B describes a procedure for finding a basis for the row space of $A$. Choice C describes the procedure for finding a basis for the null space of $A$. Choice D is a common mistake; the pivot columns of the reduced matrix RREF($A$) span Col $A$, but they are not necessarily the same vectors as those in Col $A$ itself, which is a subspace of $\mathbb{R}^4$.
4. Let $W$ be the subspace of $\mathbb{R}^4$ consisting of all vectors $\vec{x} = (x_1, x_2, x_3, x_4)$ such that $x_1 - 2x_2 + x_3 = 0$. Which of the following is a basis for $W$?
   Answer: $\{\begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\}$ The condition $x_1 - 2x_2 + x_3 = 0$ can be rewritten as $x_1 = 2x_2 - x_3$. The variables $x_2$, $x_3$, and $x_4$ are free. We can express any vector $\vec{x}$ in $W$ as: $\vec{x} = \begin{pmatrix} 2x_2 - x_3 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = x_2 \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$ The three vectors in this parametric form are linearly independent and span $W$, so they form a basis. Distractor A is the normal vector to the hyperplane, which is not in $W$. Distractor C does not span $W$. Distractor D is linearly dependent as it contains four vectors in a three-dimensional subspace.
5. Let $H$ be the subspace of $\mathbb{P}_2$ (polynomials of degree at most 2) defined by $H = \{ p(t) \in \mathbb{P}_2 \mid p(1) = 0 \}$. Which of the following is a basis for $H$?
   Answer: $\{t-1, t^2-t\}$ A polynomial $p(t) = a_2t^2 + a_1t + a_0$ is in $H$ if $p(1) = a_2 + a_1 + a_0 = 0$. This implies $a_0 = -a_1 - a_2$. Substituting back, we get $p(t) = a_2t^2 + a_1t - (a_1+a_2) = a_1(t-1) + a_2(t^2-1)$. So, $H$ is spanned by $\{t-1, t^2-1\}$. The set in choice D, $\{t-1, t^2-t\}$, is also a valid basis. Both polynomials are in $H$ since they are zero at $t=1$. They are linearly independent because neither is a scalar multiple of the other. The dimension of $H$ is 2, so this set forms a basis. Choice A is linearly dependent since $t^2-t = (t^2-1) - (t-1)$. Choice B contains $t+1$, which is not in $H$ because $1+1=2 \neq 0$. Choice C does not span $H$, as its dimension is 2.
6. Consider the subspace $T$ of $\mathbb{R}^4$ defined by $T = \{(a+b, a-b, 2a+c, b-c) : a, b, c \in \mathbb{R}\}$. Which of the following statements about a basis for $T$ is correct?
   Answer: $\{(1, 1, 2, 0), (1, -1, 0, 1), (0, 0, 1, -1)\}$ forms a basis and $\dim(T) = 3$ To find a basis for T, we express the general form: $(a+b, a-b, 2a+c, b-c) = a(1, 1, 2, 0) + b(1, -1, 0, 1) + c(0, 0, 1, -1)$. This shows that T is spanned by the three vectors $\{(1, 1, 2, 0), (1, -1, 0, 1), (0, 0, 1, -1)\}$. To verify linear independence, we check if $\alpha(1, 1, 2, 0) + \beta(1, -1, 0, 1) + \gamma(0, 0, 1, -1) = (0, 0, 0, 0)$ only when $\alpha = \beta = \gamma = 0$. This gives the system: $\alpha + \beta = 0$, $\alpha - \beta = 0$, $2\alpha + \gamma = 0$, $\beta - \gamma = 0$. From the first two equations, $\alpha = \beta = 0$, and then from the third and fourth, $\gamma = 0$. Thus the vectors are linearly independent, so option A is correct. Option B has the wrong third vector, options C and D have insufficient vectors to span the 3-dimensional space.
7. Let $U = \text{Span}\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\}$ and $V = \text{Span}\{\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\}$ be subspaces of $\mathbb{R}^3$. Which of the following is a basis for the intersection $U \cap V$?
   Answer: $\{\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\}$ A vector $\vec{x}$ is in the intersection $U \cap V$ if it can be written as a linear combination of the basis vectors for both $U$ and $V$. So, $c_1\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + c_2\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = c_3\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + c_4\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$. This gives the system $c_1 - c_3 - c_4 = 0$, $c_1 + c_2 - 2c_4 = 0$, and $c_2 + c_3 - c_4 = 0$. Solving this system, we find solutions where $c_3 = -c_4$. For a non-trivial solution, let $c_4=1$, then $c_3=-1$, $c_2=2$, $c_1=0$. This is not quite right. A simpler way is to note that the vector $\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ from the basis of $V$ can be written as $1\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + 1\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$, which means it is in $U$. Since it is also in $V$, it must be in the intersection. As the intersection is a subspace of dimension at most 1 (since the subspaces are distinct planes), this single non-zero vector forms a basis for $U \cap V$.
8. Let $S$ be the subspace of $M_{2 \times 2}$, the space of $2 \times 2$ matrices, consisting of all matrices $A$ such that $A^T = A$ (symmetric matrices). Which of the following is a basis for $S$?
   Answer: $\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\}$ A general $2 \times 2$ matrix is $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. The condition for symmetry, $A^T = A$, implies that $\begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, which requires $b=c$. Thus, any symmetric $2 \times 2$ matrix has the form $\begin{pmatrix} a & b \\ b & d \end{pmatrix}$. This can be written as a linear combination: $a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + d\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ The three matrices in this combination are linearly independent and span the space of symmetric matrices, so they form a basis. Choice A is the standard basis for all of $M_{2 \times 2}$. Choice B is a basis for skew-symmetric matrices. Choice C is linearly dependent because the last matrix is the sum of the first and third.
9. Consider the subspace $S$ of $M_{2 \times 2}(\mathbb{R})$ consisting of all $2 \times 2$ matrices $A$ such that $\text{tr}(A) = 0$ and $A^T = A$. What is the dimension of $S$, and which set forms a basis for $S$?
   Answer: Dimension 2; basis: $\left\{\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right\}$ For a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ to be in S, we need $A^T = A$ (symmetric) and $\text{tr}(A) = 0$ (traceless). Symmetry requires $b = c$, and trace zero requires $a + d = 0$, so $d = -a$. Thus $A = \begin{pmatrix} a & b \\ b & -a \end{pmatrix} = a\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. The dimension is 2, and the given matrices are linearly independent and span S. Option B has dimension 3 but the third matrix violates symmetry. Option C has only one matrix, insufficient for the 2-dimensional space. Option D's second matrix has trace 2, not 0.
10. Consider the subspace $V$ of $\mathbb{R}^6$ spanned by the vectors $u_1 = (1, 0, 1, 1, 0, 1)$, $u_2 = (0, 1, 1, 0, 1, 1)$, $u_3 = (1, 1, 2, 1, 1, 2)$, $u_4 = (2, 1, 3, 2, 1, 3)$, and $u_5 = (1, 2, 3, 1, 2, 3)$. After determining linear dependencies, which subset forms a basis for $V$?
    Answer: $\{u_1, u_2, u_5\}$ To find a basis, we need to identify which vectors are linearly independent. Notice that $u_3 = u_1 + u_2$ since $(1, 0, 1, 1, 0, 1) + (0, 1, 1, 0, 1, 1) = (1, 1, 2, 1, 1, 2) = u_3$. Also, $u_4 = 2u_1 + u_2$ since $2(1, 0, 1, 1, 0, 1) + (0, 1, 1, 0, 1, 1) = (2, 1, 3, 2, 1, 3) = u_4$. We can verify that $u_1$, $u_2$, and $u_5$ are linearly independent by checking that no nontrivial linear combination equals zero. Since $u_3$ and $u_4$ are linear combinations of $u_1$ and $u_2$, and $u_5$ is not in the span of $u_1$ and $u_2$, the vectors $\{u_1, u_2, u_5\}$ form a basis for the 3-dimensional subspace V. The other options either include dependent vectors or cannot span the full space.
11. Let $\vec{v}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$, $\vec{v}_2 = \begin{pmatrix} -2 \\ 3 \\ -1 \end{pmatrix}$, and $\vec{v}_3 = \begin{pmatrix} 1 \\ 0 \\ k \end{pmatrix}$. For which value of $k$ does the set $\{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$ NOT form a basis for $\mathbb{R}^3$?
    Answer: $k = 5$ The set will not form a basis for $\mathbb{R}^3$ if the vectors are linearly dependent. This occurs if the matrix formed by these vectors has a determinant of zero. $A = \begin{pmatrix} 1 & -2 & 1 \\ -1 & 3 & 0 \\ 2 & -1 & k \end{pmatrix}$. The determinant is $1(3k - 0) - (-2)(-k - 0) + 1(1 - 6) = 3k - 2k - 5 = k - 5$. The set is linearly dependent when the determinant is zero, which means $k - 5 = 0$, or $k=5$. For any other value of $k$, the determinant is non-zero, and the vectors form a basis for $\mathbb{R}^3$.
12. Let $S = \{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$ be a basis for a vector space $W$. Which of the following statements must be true?
    Answer: The set $\{\vec{v}_1, \vec{v}_2, 2\vec{v}_3\}$ is another basis for $W$. A basis must be linearly independent and span the space. Choice C is correct because replacing a basis vector with a non-zero scalar multiple of itself results in a new set that is still linearly independent and spans the same space. Choice A is false because $\{\vec{v}_1, \vec{v}_2\}$ is linearly independent but cannot span the three-dimensional space $W$. Choice B is false because the set is linearly dependent; the fourth vector is a linear combination of the first and third. Choice D is false; for example, the set $\{\vec{v}_1, 2\vec{v}_1\}$ consists of two vectors from $W$ but is linearly dependent.
13. Let $W$ be the subspace of $M_{3 \times 3}(\mathbb{R})$ consisting of all matrices $A$ such that $A + A^T = 2I$, where $I$ is the $3 \times 3$ identity matrix. What is the dimension of $W$?
    Answer: $\dim(W) = 3$ For a matrix $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ to satisfy $A + A^T = 2I$, we need $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} + \begin{pmatrix} a & d & g \\ b & e & h \\ c & f & i \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}$. This gives us: $2a = 2$ (so $a = 1$), $2e = 2$ (so $e = 1$), $2i = 2$ (so $i = 1$), $b + d = 0$ (so $d = -b$), $c + g = 0$ (so $g = -c$), and $f + h = 0$ (so $h = -f$). Therefore, $A = \begin{pmatrix} 1 & b & c \\ -b & 1 & f \\ -c & -f & 1 \end{pmatrix}$. The matrix depends on 3 free parameters ($b$, $c$, $f$), so $\dim(W) = 3$.
14. Let $W$ be the subspace of $\mathbb{R}^4$ spanned by the vectors $\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}$, $\vec{v}_2 = \begin{pmatrix} -2 \\ -4 \\ 1 \\ 5 \end{pmatrix}$, and $\vec{v}_3 = \begin{pmatrix} 3 \\ 6 \\ -1 \\ -6 \end{pmatrix}$. Which of the following sets is a basis for $W$?
    Answer: $\{\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} -2 \\ -4 \\ 1 \\ 5 \end{pmatrix}\}$ To find a basis for the subspace spanned by a set of vectors, we form a matrix with these vectors as columns and row-reduce it to find the pivot columns. The corresponding original vectors form the basis. Let $A = [\vec{v}_1 \; \vec{v}_2 \; \vec{v}_3] = \begin{pmatrix} 1 & -2 & 3 \\ 2 & -4 & 6 \\ 0 & 1 & -1 \\ -1 & 5 & -6 \end{pmatrix}$. Row reduction proceeds as: $\begin{pmatrix} 1 & -2 & 3 \\ 0 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 3 & -3 \end{pmatrix} \to \begin{pmatrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$. The pivot columns are the first and second columns. Therefore, a basis for $W$ is given by the first two original vectors, $\{\vec{v}_1, \vec{v}_2\}$.
15. Which of the following sets forms a basis for the null space of the matrix $A = \begin{pmatrix} 1 & -2 & 0 & 5 \\ -1 & 2 & 1 & -8 \end{pmatrix}$?
    Answer: $\{\begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -5 \\ 0 \\ 3 \\ 1 \end{pmatrix}\}$ To find a basis for the null space of $A$, we solve the homogeneous system $A\vec{x} = \vec{0}$. The augmented matrix is $\begin{pmatrix} 1 & -2 & 0 & 5 & 0 \\ -1 & 2 & 1 & -8 & 0 \end{pmatrix}$. Adding the first row to the second gives $\begin{pmatrix} 1 & -2 & 0 & 5 & 0 \\ 0 & 0 & 1 & -3 & 0 \end{pmatrix}$. This is the reduced row echelon form. The corresponding equations are $x_1 - 2x_2 + 5x_4 = 0$ and $x_3 - 3x_4 = 0$. The free variables are $x_2$ and $x_4$. We can write the solution as $x_1 = 2x_2 - 5x_4$ and $x_3 = 3x_4$. In parametric vector form, this is $\vec{x} = x_2 \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} -5 \\ 0 \\ 3 \\ 1 \end{pmatrix}$. The vectors in this form constitute a basis for Nul $A$.
16. Let $W$ be the subspace of $\mathbb{R}^3$ spanned by the set $\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}\}$. Which of the following sets is a basis for $W$?
    Answer: $\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\}$ The given set of four vectors spans $W$. To find a basis, we must find a linearly independent subset that still spans $W$. Let the vectors be $\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4$. By inspection, $\vec{v}_3 = \vec{v}_1 + \vec{v}_2$ and $\vec{v}_4 = 2\vec{v}_1$. Thus, $\vec{v}_3$ and $\vec{v}_4$ are redundant and can be removed. The remaining set is $\{\vec{v}_1, \vec{v}_2\} = \{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\}$. These two vectors are not scalar multiples of each other, so they are linearly independent. Therefore, they form a basis for $W$. Choice A is linearly dependent. Choice B is linearly dependent. Choice D is a vector that is not in the subspace $W$.

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